Answer:
a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
b) T = 295.37 K
Explanation:
Given;
Initial temperature of tea T1 = 31 C
Initial temperature of ice T2 = 0 C
Mass of tea m1 = 0.89 kg
Mass of ice m2 = 0.075kg
The heat capacity of both water and tea c = 4186 J/(kg⋅K)
the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg
And T = the final temperature of the mixture
Heat loss by tea = heat gained by ice
m1c∆T1 = m2c∆T2 + m2Lf
m1c(T1-T) = m2c(T-T2) + m2Lf
m1cT1 - m1cT = m2cT - m2cT2 + m2Lf
m1cT + m2cT = m1cT1 + m2cT2 - m2Lf
T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
Substituting the values;
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)
T = 22.37 °C
T = 273 + 22.37 K
T = 295.37 K
Answer:
The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.
Explanation:
Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.
Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.
Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.
It i believe it would be 7.3 × 10∧3.
correct me if im wrong
The first law of thermodynamics says that the variation of internal energy of a system is given by:

where Q is the heat delivered by the system, while W is the work done on the system.
We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system
So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J
So, the variation of internal energy of the system is